Alternative definition

You might also find it helpful to know that you can characterize closed a set by knowledge of the limits of its sequences. Let $\displaystyle (X,d)$ be a metric space, then $\displaystyle F\subset X$ is closed if and only if, whenever the sequence $\displaystyle (x_{n})_{n\in\mathbb{N}} \in X$ converges - to $\displaystyle x,$ say, - then it follows that $\displaystyle x\in F$. I.e.: $\displaystyle F$ is closed if and only if it contains all of its limits. You can prove this using your definition.

So if F is not open, I assume F is closed. What is F^c? It is U. So U is open.

Be very careful with statements such as that.
A door is either open or closed. That ain't true for sets.
The set $\displaystyle [0,1)$ is neither open nor closed.
Its complement $\displaystyle (-\infty,0)\cup [1,\infty)$ is also neither open nor closed

If is true that a set is closed in and only if its complement is open.

As has already been pointed out, the standard metric space definition of closed sets is:A set is closed if and only if it contains all of its limit points.

closed set-biconditional statement

Originally Posted by Plato

Be very careful with statements such as that.
A door is either open or closed. That ain't true for sets.
The set $\displaystyle [0,1)$ is neither open nor closed.
Its complement $\displaystyle (-\infty,0)\cup [1,\infty)$ is also neither open nor closed

If is true that a set is closed in and only if its complement is open.

As has already been pointed out, the standard metric space definition of closed sets is:A set is closed if and only if it contains all of its limit points.

This makes sense to me. So, the definition is a biconditional statement rather than an implication. I just wanted to make sure so that I can use is properly in a proof. Thank you so much, Plato.